Simplify the following expression and state the condition under which the simplification is valid. $x = \dfrac{-3n^2 + 3}{-7n^2 + 56n + 63}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ x = \dfrac {-3(n^2 - 1)} {-7(n^2 - 8n - 9)} $ $ x = \dfrac{3}{7} \cdot \dfrac{n^2 - 1}{n^2 - 8n - 9} $ Next factor the numerator and denominator. $ x = \dfrac{3}{7} \cdot \dfrac{(n + 1)(n - 1)}{(n + 1)(n - 9)}$ Assuming $n \neq -1$ , we can cancel the $n + 1$ $ x = \dfrac{3}{7} \cdot \dfrac{n - 1}{n - 9}$ Therefore: $ x = \dfrac{ 3(n - 1)}{ 7(n - 9)}$, $n \neq -1$